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Calculate $\Delta H^{\circ}$ for the reaction: $Na_2O+SO_3 \longrightarrow Na_2SO_4$ given the following :

\[\begin {array} {1 1} (A)\;Na+H_2O \longrightarrow NaOH+\frac{1}{2} H_2 \qquad \Delta H^{\circ}=-146\;kJ \\(B)\;Na_2SO_4+H_2O \longrightarrow 2NaOH+ SO_3 \qquad \Delta H^{\circ}=+418\;kJ \\ (C)\;2Na_2O+2H_2 \longrightarrow 4Na+2 H_2O \qquad \Delta H^{\circ}=+259\;kJ \end {array}\] \[\begin {array} {1 1} (1)\;+823\;kJ & \quad (2)\;-581\;kJ \\ (3)\;-435\;kJ & \quad (4)\;+531\;kJ \end {array}\]

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(2) -581 kJ
answered Nov 7, 2013 by pady_1

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